Discussion
The angle of elevation of the top of a tower from a point on the ground is 30° . On walking 100m towards the tower the angle of elevation changes to 60° . Find the height of the tower.
Answer: D
Let the tower be \( h \). The two positions give:
\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{60} = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}} \]
The 100 m walked is the difference:
\[ h\sqrt{3} - \frac{h}{\sqrt{3}} = 100 \quad\Longrightarrow\quad \frac{3h - h}{\sqrt{3}} = 100 \]
\[ \frac{2h}{\sqrt{3}} = 100 \quad\Longrightarrow\quad h = 50\sqrt{3} \approx 86.6 \text{ m} \]
The pairing \( 30^{\circ}/60^{\circ} \) is convenient because \( \sqrt{3} - \tfrac{1}{\sqrt{3}} \) collapses to \( \tfrac{2}{\sqrt{3}} \).
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